Using symbolically the coordinates of an intersection point between curves for filling the area below a rectangle

Question

I would like to fill the area under the rectangles that have the origin and the working point (Q_1, H_1) and (Q_2, H_2), respectively, obtained as intersections between the two red curves with the blue one. I managed to pinpoint the intersection point and to draw the segments defining the rectangle, and then I tried to use fill between but the output was coloured trapezoid between the segment and the whole horizontal axis. I started thinking the missing element is the right end-point of the interval over which the horizontal axis path is defined (which has to be Q_1 and not xmax), so I looked for a solution to this problem, without any relevant success. I am not very familiar with this package (that is the reason why maybe this question is almost a duplicate but I cannot determine it), so I have not been able to use properly let or pgfextractx functions for this purpose. How can I reference symbolically the abscissa of the intersection point (OP1) and use it for defining properly the useful interval for fill between?

Here follows my attempt:

\usepackage{tikz}
\usepackage{pgfplots}
\pgfplotsset{compat=1.14}
\usetikzlibrary{arrows.meta,babel,calc,intersections}
\usepgfplotslibrary{fillbetween}

\begin{figure}[H] %% figure is closed
\begin{center}
\begin{tikzpicture}
\begin{axis}[
xmin=0, xmax=1, ymin=0, ymax=1.5,
axis x line=middle,
axis y line=middle,
xlabel=$Q$,ylabel=$H$,
ticks=none,
]
\addplot[name path =pump,blue,domain=0:1] {-0.5*x^2+1};
\addplot[red,domain=0:1,name path=load1] {0.5*x^2+0.4*x+0.5};
\addplot[red,domain=0:1,name path=load2] {2*x^2+1.6*x+0.5};

\path [name intersections={of=load1 and pump}]; 
\coordinate [label= ${(Q_1,H_1)}$ ] (OP1) at (intersection-1);
\path [name intersections={of=load2 and pump}]; 
\coordinate [label= ${(Q_2,H_2)}$ ] (OP2) at (intersection-1);

\draw[name path=opv1] (OP1) -- (OP1|-0,0);
\draw[name path=oph1] (OP1) -- (0,0 |- OP1);
\draw[name path=opv2] (OP2) -- (OP2|-0,0);
\draw[name path=oph2] (OP2) -- (0,0 |- OP2);

\path[name path=zero]
(\pgfkeysvalueof{/pgfplots/xmin},0) --
(\pgfkeysvalueof{/pgfplots/xmax},0);

\addplot[orange]fill between[of=oph1 and zero];
\end{axis}

\foreach \point in {OP1,OP2}
\fill [red] (\point) circle (2pt);


\end{tikzpicture}
\end{center}
\caption{System working point}
\end{figure}

Answer 1

So if all you need is two rectangles, then fill between is kind of an overcomplication, you can just use \fill (x,y) rectangle (u,v);. You already have the coordinates. Below I also added the backgrounds library, and placed the rectangles inside a scope environment with [on background layer], so that the rectangles are placed behind the plot lines.

I used \filldraw, but if you don't want the border drawn, change that to \fill, and remove the draw=black option.

\documentclass[border=5mm]{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.14}
\usetikzlibrary{arrows.meta,babel,calc,intersections,backgrounds}
\usepgfplotslibrary{fillbetween}
\begin{document}
\begin{tikzpicture}
\begin{axis}[
xmin=0, xmax=1, ymin=0, ymax=1.5,
axis x line=middle,
axis y line=middle,
xlabel=$Q$,ylabel=$H$,
ticks=none,
]
\addplot[name path =pump,blue,domain=0:1] {-0.5*x^2+1};
\addplot[red,domain=0:1,name path=load1] {0.5*x^2+0.4*x+0.5};
\addplot[red,domain=0:1,name path=load2] {2*x^2+1.6*x+0.5};

\path [name intersections={of=load1 and pump}]; 
\coordinate [label= ${(Q_1,H_1)}$ ] (OP1) at (intersection-1);
\path [name intersections={of=load2 and pump}]; 
\coordinate [label= ${(Q_2,H_2)}$ ] (OP2) at (intersection-1);

\begin{scope}[on background layer]
\filldraw[fill=orange!50,draw=black] (0,0) rectangle node {foo} (OP2);
\filldraw[fill=blue!80!red!50!white,draw=black] (0,0-|OP2) rectangle node {bar} (OP1);
\end{scope}
\end{axis}

\foreach \point in {OP1,OP2}
  \fill [red] (\point) circle (2pt);

\end{tikzpicture}
\end{document}

Answer 2

As Torbjørn T. and Gernot already stated, I am also not 100% sure what you really want to achieve. But I guess the answer of Torbjørn T. is pretty much what you are searching for.

So this is pretty much the same answer, but using the fill between features.

% used PGFPlots v1.14
\documentclass[border=5pt]{standalone}
\usepackage{pgfplots}
    \usetikzlibrary{
        intersections,
        pgfplots.fillbetween,
    }
    \pgfplotsset{compat=1.11}
\begin{document}
\begin{tikzpicture}
    \begin{axis}[
        xmin=0, xmax=1, ymin=0, ymax=1.5,
        axis x line=middle,
        axis y line=middle,
        xlabel=$Q$,ylabel=$H$,
        ticks=none,
    ]
        \addplot[blue,domain=0:1,name path=pump] {-0.5*x^2+1};
        \addplot[red,domain=0:1,name path=load1] {0.5*x^2+0.4*x+0.5};
        \addplot[red,domain=0:1,name path=load2] {2*x^2+1.6*x+0.5};

        \path [name intersections={of=load1 and pump}];
            \coordinate [label= ${(Q_1,H_1)}$ ] (OP1) at (intersection-1);
        \path [name intersections={of=load2 and pump}];
            \coordinate [label= ${(Q_2,H_2)}$ ] (OP2) at (intersection-1);

        \draw [name path=opv1] (OP1) -- (OP1 |- 0,0);
        \draw [name path=oph1] (OP1) -- (0,0 |- OP1);
        \draw [name path=opv2] (OP2) -- (OP2 |- 0,0);
        \draw [name path=oph2] (OP2) -- (0,0 |- OP2);

        \path [name path=zero]
            (\pgfkeysvalueof{/pgfplots/xmin},0) --
            (\pgfkeysvalueof{/pgfplots/xmax},0);

        \addplot [orange] fill between [
            of=oph1 and zero,
            % -----------------------------------------------------------------
            % in order to achieve the desired result you can add a `soft clip'
            % path which cuts off the unwanted rest not inside of this `soft
            % clip' path and you need (in this case) also add `reverse=true'
            % explicitly, otherwise you get another unwanted result
            reverse=true,
            soft clip={
                % (depending on what you exactly need use one the following
                % starting corrdinates)
%                (OP2 |- 0,\pgfkeysvalueof{/pgfplots/ymin})
                (\pgfkeysvalueof{/pgfplots/xmin},\pgfkeysvalueof{/pgfplots/ymin})
                    rectangle
                (OP1)
            },
            % -----------------------------------------------------------------
        ];

        % add some text centered in the rectangle (as requested in the comments)
        \path
%            % (to show how it is drawn ...)
%            [draw,dashed]
                (\pgfkeysvalueof{/pgfplots/xmin},\pgfkeysvalueof{/pgfplots/ymin})
                    -- node [pos=0.5] {some text} (OP1);
        ;

    \end{axis}

    \foreach \point in {OP1,OP2} {
        \fill [red] (\point) circle (2pt);
    }

\end{tikzpicture}
\end{document}

Answer 3

A PSTricks solution using the pst-plot package:

\documentclass{article}

\usepackage{pst-plot}
\usepackage{xfp}

% The blue-coloured graph.
\def\aA{-0.5}
\def\bA{0}
\def\cA{1}
\def\fA(#1){(\aA*(#1)^2+\bA*(#1)+\cA)}
% The "first" red-coloured graph.
\def\aB{2}
\def\bB{1.6}
\def\cB{0.5}
\def\fB(#1){(\aB*(#1)^2+\bB*(#1)+\cB)}
% The "second" red-coloured graph.
\def\aC{0.5}
\def\bC{0.4}
\def\cC{0.5}
\def\fC(#1){(\aC*(#1)^2+\bC*(#1)+\cC)}
% Intersection points, x-coordinates.
\def\xAB{\fpeval{(-(\bA-\bB)-sqrt((\bA-\bB)^2-4*(\aA-\aB)*(\cA-\cB)))/(2*(\aA-\aB))}}
\def\xAC{\fpeval{(-(\bA-\bC)-sqrt((\bA-\bC)^2-4*(\aA-\aC)*(\cA-\cC)))/(2*(\aA-\aC))}}

\begin{document}

{\psset{
   xunit = 6,
   yunit = 3,
   dimen = m,
   algebraic
 }
\begin{pspicture}(1.2,1.5)
 {\psset{fillstyle = solid}
  \psframe[
    fillcolor = orange!60
  ](0,0)(\xAB,\fpeval{\fA(\xAB)})
  \rput(\fpeval{0.5*\xAB},\fpeval{0.5*\fA(\xAB)}){cat}
  \psframe[
    fillcolor = blue!60
  ](\xAB,0)(\xAC,\fpeval{\fA(\xAC)})
  \rput(\fpeval{0.5*(\xAC+\xAB)},\fpeval{0.5*\fA(\xAC)}){dog}}
  \psaxes[
    ticks = none,
    labels = none
  ]{->}(0,0)(-0.05,-0.1)(0.8,1.5)[$Q$,120][$H$,330]
  \psplot[
    linecolor = blue
  ]{0}{0.8}{\fA(x)}
  \psplot[
    linecolor = red
  ]{0}{0.4}{\fB(x)}
  \psplot[
    linecolor = red
  ]{0}{0.8}{\fC(x)}
  \psdots[
    dotstyle = o,
    fillcolor =red
  ](\xAB,\fpeval{\fA(\xAB)})(\xAC,\fpeval{\fA(\xAC)})
\end{pspicture}}

\end{document}

All you have to do is change the values of the coefficients of the three second degree polynomials and the drawing will be adjusted accordingly. (You have to change the plot ranges and axes ranges manually, if needed.)

Addition

If you add

\uput[90](\xAB,\fpeval{\fA(\xAB)}){$(G_{1},H_{1})$}
\uput[90](\xAC,\fpeval{\fA(\xAC)}){$(G_{2},H_{2})$}

to the drawing, you will get the labels of the two intersection points.